## College Algebra (11th Edition)

$$\frac{-2}{x-3}+\frac{3}{x+3}=\frac{-12}{(x+3)(x-3)}$$ $$(x+3)(x-3)(\frac{-2}{x-3}+\frac{3}{x+3})=(x+3)(x-3)(\frac{-12}{(x+3)(x-3)})$$ $$-2(x+3)+3(x-3)=-12$$ $$-2x-6+3x-9=-12$$ $$x-15=-12$$ $$x=3$$ However, 3 is an extraneous solution. There is no solution.